Problem # 46 Exercises [Solved]
Attention: These exercises apply the principles and theory explained in Issue # 41(Rates of Radioactive decay, Half-life time, Dating).
1/ ln N/No = ln (4min-1/16min-1) = ln 0.25 = -kt
t1/2 = 0.693/k = 5,600 years
k = 0.693/5,600 y = 1.2x10-4 y-1
t= (ln 0.25)/-1.2x10-4 y-1 = 11,552 years = the age of the wood.
2/ ln N/No = ln (0.953/1000) = -kt = -kx2 y
ln 9.53 x 10-2 = -0.04814 = -k x 2 y
k = 0.04814/2y = 0.02407 y-1
(a) t1/2 = 0.693/0.02407 y-1 = 28.8 years = half-life time of Sr-90.
(b) N = Noe-kt = 1g x e-0.02407x5 = 0.887 g = the remaining Sr-90 after 5 years.
3/ There is no effect on chemical properties since the atomic number and the number of electrons remain unchanged. But an isotope N-15 with slightly different physical properties than N-14 will be generated.